It is easy to see that “\(\supset \)” holds. As for “\(\subset \)”, suppose \(E\) is a finite unramified abelian \(p\)-extension of \(K_\infty \), we want to prove \(E\subset L_\infty \). The proof consists of the following steps:

(1) There exists an integer \(n_0\geq 0\) such that \(E/K_{n_0}\) is Galois.

(2) There exists an integer \(n_1\geq n_0\) such that \({\mathrm{Gal}}(E/K_{n_1})\) is abelian. So \({\mathrm{Gal}}(E/K_{n_1})\cong {\mathrm{Gal}}(K_\infty /K_{n_1})\times G\), where \(G\) is a finite abelian group, corresponding to some \(E_{n_1}/K_{n_1}\) finite abelian extension, so that \(E=K_\infty E_{n_1}\).

(3) There exists an integer \(n_2\geq n_1\) such that \(E_{n_1}K_{n_2}/K_{n_2}\) is a finite unramified abelian \(p\)-extension. Therefore \(E_{n_1}K_{n_2}\subset L_{n_2}\), hence \(E\subset L_{n_2}K_\infty \subset L_\infty \).

Let \(D_{\mathfrak {l}}\) be the decomposition subgroup of \({\mathfrak {l}}\) in \(\Gamma :={\mathrm{Gal}}(K_\infty /K)\). Let \(l=\operatorname{char}({\mathcal{O}}_K/{\mathfrak {l}})\neq p\), then \(K_{\mathfrak {l}}/{\mathbb {Q}}_l\) is a finite extension, and let \((K_\infty )_{\mathfrak {l}}:=\varinjlim (K_n)_{\mathfrak {l}}\), then \(D_{\mathfrak {l}}={\mathrm{Gal}}((K_\infty )_{\mathfrak {l}}/K_{\mathfrak {l}})\) is a subgroup of \({\mathbb {Z}}_p\).

We have \(K_{\mathfrak {l}}\subset K_{\mathfrak {l}}^{\mathrm{unr}}\subset K_{\mathfrak {l}}^{\mathrm{ab}}\), and

So \(K_{\mathfrak {l}}\) has at least one \({\mathbb {Z}}_p\)-extension, i.e. the unique unramified \({\mathbb {Z}}_p\)-extension. If there are other \({\mathbb {Z}}_p\)-extensions of \(K_{\mathfrak {l}}\), then there exists a Galois extension of \(K_{\mathfrak {l}}\) with Galois group isomorphic to \({\mathbb {Z}}_p^2\).

However, \({\mathrm{Gal}}(K_{\mathfrak {l}}^{\mathrm{ab}}/K_{\mathfrak {l}}^{\mathrm{unr}})\) doesn’t have a quotient isomorphic to \({\mathbb {Z}}_p\), because by local class field theory, \({\mathrm{Gal}}(K_{\mathfrak {l}}^{\mathrm{ab}}/K_{\mathfrak {l}}^{\mathrm{unr}})\cong {\mathcal{O}}_{K_{\mathfrak {l}}}^\times \cong (\text{a finite group})\times {\mathbb {Z}}_l^{[K_{\mathfrak {l}}:{\mathbb {Q}}_l]}\) which is a finite group times a pro-\(l\) group, obviously it doesn’t have a quotient isomorphic to \({\mathbb {Z}}_p\). So there is only one \({\mathbb {Z}}_p\)-extension of \(K_{\mathfrak {l}}\), note that \((K_\infty )_{\mathfrak {l}}/K_{\mathfrak {l}}\) is either trivial or a \({\mathbb {Z}}_p\)-extension, in both cases it must be contained in \(K_{\mathfrak {l}}^{\mathrm{unr}}\).

Consider \({\mathrm{Gal}}(K_\infty /K_n)=\Gamma ^{p^n}\) with topological generator \(\gamma ^{p^n}\). Let \(E_n\) be the maximal abelian extension of \(K_n\) contained in \(L_\infty \), so that \({\mathrm{Gal}}(L_\infty /E_n)={\mathrm{Gal}}(L_\infty /K_n)' =(\gamma ^{p^n}-1)X_\infty \). Note that \(X_\infty /TX_\infty \cong {\mathrm{Gal}}(E_0/K_\infty )\) is a finitely generated \({\mathbb {Z}}_p\)-module, so by Nakayama lemma (Lemma 2.5) we know that \(X_\infty \) is finitely generated \(\Lambda \)-module.

Recall that if \(r=\operatorname{rank}_\Lambda X_\infty \), then \(\operatorname{rank}_{{\mathbb {Z}}_p}\left(X_\infty /(\gamma ^{p^n}-1)X_\infty \right)=rp^n+O(1)\) as \(n\to \infty \) (Lemma 5.4). So in order to prove \(X_\infty \) is torsion (i.e. \(r=0\)), we only need to prove \(\operatorname{rank}_{{\mathbb {Z}}_p}\left(X_\infty /(\gamma ^{p^n}-1)X_\infty \right)\) is bounded.

Let \({\mathfrak {p}}_1,\cdots ,{\mathfrak {p}}_t\) be the primes of \(K\) which are ramified in \(K_\infty /K\). Then \(t\) is finite (by Proposition 5.5), and number of primes of \(K_\infty \) lying over \({\mathfrak {p}}_1,\cdots ,{\mathfrak {p}}_t\) is also finite, because for each \(i\), the index \([\Gamma :D_{{\mathfrak {p}}_i}]\) is finite. So let \(s_n\) be the number of primes of \(K_n\) which are ramified in \(K_\infty /K_n\), then \(s_n\) is bounded.

Consider \(E_n/K_n\). Recall that \(L_n\) is the \(p\)-Hilbert class field of \(K_n\). Then \(L_n\subset L_\infty \), \(K_\infty \subset L_\infty \), so \(L_nK_\infty \subset E_n\). Let \(I_1,\cdots ,I_{s_n}\) be the inertia subgroups of \({\mathrm{Gal}}(E_n/K_n)\) for the ramified primes. For \(1\leq j\leq s_n\), \(I_j\cap {\mathrm{Gal}}(E_n/K_\infty )=\{ 1\} \), because \(E_n/K_\infty \) is unramified. Therefore \(I_j\) maps injectively to a closed subgroup of \(\Gamma ^{p^n}\) via the map \({\mathrm{Gal}}(E_n/K_n)\to {\mathrm{Gal}}(K_\infty /K_n)\), in particular, \(I_j\) is isomorphic to \({\mathbb {Z}}_p\).

Now we come to the key point. Let \(I:=I_1\cdots I_{s_n}\subset {\mathrm{Gal}}(E_n/K_n)\), then \(\operatorname{rank}_{{\mathbb {Z}}_p}I\leq s_n\), and \(E_n^I\) is the maximal unramified abelian pro-\(p\) extension of \(K_n\), so \(E_n^I=L_n\), hence \(\operatorname{rank}_{{\mathbb {Z}}_p}{\mathrm{Gal}}(E_n/K_n)\leq s_n\), because \({\mathrm{Gal}}(L_n/K_n)\cong {\mathrm{Cl}}(K_n)(p)\) is a finite group. Therefore \(\operatorname{rank}_{{\mathbb {Z}}_p}{\mathrm{Gal}}(E_n/K_\infty )\leq s_n-1\), because \({\mathrm{Gal}}(K_\infty /K_n)=\Gamma ^{p^n}\) is of \({\mathbb {Z}}_p\)-rank \(1\). Note that \({\mathrm{Gal}}(E_n/K_\infty )\cong X_\infty /(\gamma ^{p^n}-1)X_\infty \), so \(\operatorname{rank}_{{\mathbb {Z}}_p}\left(X_\infty /(\gamma ^{p^n}-1)X_\infty \right)\) is bounded.

Let \(G:={\mathrm{Gal}}(L_\infty /K)\). We claim that \((\gamma -1)X_\infty \) is a closed normal subgroup of \(G\) and \(X_\infty /(\gamma -1)X_\infty \cong X_0={\mathrm{Cl}}(K)(p)\). We have a group extension

which induces

so \(G/(\gamma -1)X_\infty \) is abelian, and \((\gamma -1)X_\infty \) is a closed normal subgroup. The proof of Proposition 5.8 can be easily modified to show that \((\gamma -1)X_\infty =G'\). Let \(E=L_\infty ^{G'}\) be the maximal abelian extension of \(K\) contained in \(L_\infty \). Let \(I_{\mathfrak {P}}\) be the inertia subgroup of \({\mathrm{Gal}}(E/K)\) for a prime \({\mathfrak {P}}\mid {\mathfrak {p}}\), then \(L_0=E^{I_{\mathfrak {P}}}\). Let \(H={\mathrm{Gal}}(E/K_\infty )\) (so that \(K_\infty =E^H\)), then \(L_0\cap K_\infty =K\), \(H\cap I_{\mathfrak {P}}=\{ 1\} \), so \(E=L_0K_\infty \), and \({\mathrm{Gal}}(E/K_\infty )\xrightarrow \sim {\mathrm{Gal}}(L_0/K)\) is a natural isomorphism. Hence we conclude that \(X_\infty /(\gamma -1)X_\infty \xrightarrow \sim X_0\).

In general, consider \(L_n/K_n\), the \(p\)-Hilbert class field of \(K_n\), so that \({\mathrm{Gal}}(L_n/K_n)=:X_n\cong {\mathrm{Cl}}(K_n)(p)\). We have \({\mathrm{Gal}}(K_\infty /K_n)\cong \Gamma ^{p^n}\cong p^n{\mathbb {Z}}_p\) with topological generator \(\gamma ^{p^n}\). We have \(K_\infty /K_n\) is ramified at only one place and is totally ramified. So similarly, we get \(X_\infty /(\gamma ^{p^n}-1)X_\infty \xrightarrow \sim X_n\).

“\(\supset \)” is trivial. As for “\(\subset \)”, we have an exact sequence

and \({\mathcal{G}}/(g-1){\mathcal{X}}\) is a central extension of \({\mathcal{G}}/{\mathcal{X}}\) (which is cyclic) by \({\mathcal{X}}/{(g-1){\mathcal{X}}}\), so it is abelian, so \({\mathcal{G}}'\subset (g-1){\mathcal{X}}\).